Problem: Let $g(x)=\dfrac{2x+5}{x^2-3}$. Find $g'(-2)$. Choose 1 answer: Choose 1 answer: (Choice A) A $-6$ (Choice B) B $9$ (Choice C) C $6$ (Choice D) D $\dfrac19$
Answer: Let's first find the derivative of $g$, i.e. the expression for $g'(x)$. Then, we can plug $x=-2$, into $g'(x)$ and evaluate. $g$ is a rational function. To find the derivative of rational functions, we use the quotient rule : $\begin{aligned} \dfrac{d}{dx}\left[\dfrac{u(x)}{v(x)}\right]&=\dfrac{\dfrac{d}{dx}[u(x)]v(x)-u(x)\dfrac{d}{dx}[v(x)]}{[v(x)]^2} \\\\ &=\dfrac{u'(x)v(x)-u(x)v'(x)}{[v(x)]^2} \end{aligned}$ Let's differentiate! = g ′ ( x ) = d d x ( 2 x + 5 x 2 − 3 ) = ( x 2 − 3 ) d d x ( 2 x + 5 ) − ( 2 x + 5 ) d d x ( x 2 − 3 ) ( x 2 − 3 ) 2 = ( x 2 − 3 ) ( 2 ) − ( 2 x + 5 ) ( 2 x ) ( x 2 − 3 ) 2 = 2 x 2 − 6 − 4 x 2 − 10 x ( x 2 − 3 ) 2 = − 2 x 2 − 10 x − 6 ( x 2 − 3 ) 2 The quotient rule Differentiate ( 2 x + 5 ) & ( x 2 − 3 ) Expand \begin{aligned} &\phantom{=}g'(x) \\\\ &=\dfrac{d}{dx}\left(\dfrac{2x+5}{x^2-3}\right) \\\\ &=\dfrac{(x^2-3)\dfrac{d}{dx}(2x+5)-(2x+5)\dfrac{d}{dx}(x^2-3)}{(x^2-3)^2}&&\gray{\text{The quotient rule}} \\\\ &=\dfrac{(x^2-3)(2)-(2x+5)(2x)}{(x^2-3)^2}&&\gray{\text{Differentiate }(2x+5)\text{ & }(x^2-3)} \\\\ &=\dfrac{2x^2-6-4x^2-10x}{(x^2-3)^2}&&\gray{\text{Expand}} \\\\ &=\dfrac{-2x^2-10x-6}{(x^2-3)^2} \end{aligned} So we found that $g'(x)=\dfrac{-2x^2-10x-6}{(x^2-3)^2}$. Now let's plug $x= {-2}$ into $g'$ : $\begin{aligned} &\phantom{=}g'( {-2}) \\\\ &=\dfrac{-2( {-2})^2-10({ -2})-6}{(( {-2})^2-3)^2} \\\\ &=\dfrac{-8+20-6}{(1)^2} \\\\ &=6 \end{aligned}$ In conclusion, $g'(-2)=6$.